لمّاح✨ PMAU
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لمّاح✨ PMAU Ingliz til segmentidagi kanali faol ishtirokchi. Hozirda hamjamiyat 11 174 obunachidan iborat bo'lib, Taʼlim toifasida 17 557-o'rinni va AQSH mintaqasida 3 132-o'rinni egallagan.
📊 Auditoriya ko‘rsatkichlari va dinamika
невідомо sanasidan buyon loyiha tez o‘sib, 11 174 obunachiga ega bo‘ldi.
05 Oktabr, 2026 dagi oxirgi ma’lumotlarga ko‘ra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni -224 ga, so‘nggi 24 soatda esa -9 ga o‘zgardi va umumiy qamrov yuqori darajada qolmoqda.
- Tasdiqlash holati: Tasdiqlanmagan
- Jalb etish (ER): Auditoriya o‘rtacha 11.27% darajada jalb etiladi. Nashrdan keyingi dastlabki 24 soatda kontent odatda umumiy obunachilar sonining 3.41% ini tashkil etuvchi reaksiyalarni to‘playdi.
- Post qamrovi: Har bir post o‘rtacha 1 261 marta ko‘riladi; birinchi sutkada odatda 381 ta ko‘rish yig‘iladi.
- Reaksiyalar va o‘zaro ta’sir: Auditoriya faol: har bir postga o‘rtacha 0 ta reaksiya keladi.
- Tematik yo‘nalishlar: Kontent قَنَاة, مَادَّة, كُلِّيَّة, مُتَطَوِّع, مَسَاء kabi asosiy mavzularga jamlangan.
📝 Tavsif va kontent siyosati
Kanal uchun tavsif kiritilmagan.
Yuqori yangilanish chastotasi (oxirgi ma’lumot 06 Oktabr, 2026 da olingan) sababli kanal doimo dolzarb va katta qamrovli bo‘lib qoladi. Analitika auditoriya kontent bilan faol hamkorlik qilishini, uni Taʼlim toifasidagi muhim ta’sir nuqtasiga aylantirishini ko‘rsatadi.
11 174
Obunachilar
-924 soatlar
-377 kun
-22430 kun
Ma'lumot yuklanmoqda...
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| Sana | Obunachilarni jalb qilish | Esdaliklar | Kanallar | |
| 06 Oktabr | 0 | |||
| 05 Oktabr | 0 | |||
| 04 Oktabr | 0 | |||
| 03 Oktabr | 0 | |||
| 02 Oktabr | 0 | |||
| 01 Oktabr | 0 |
Kanal postlari
| 2 | ما هو مستواك الدراسي الحالي | 626 |
| 3 | ماهو مستواك الدراسي الحالي ⁉️ | 13 |
| 4 | تحضيري خطة A | 1 |
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| 6 |  Matn yo'q... | 1 532 |
| 7 |  Matn yo'q... | 1 |
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متابعة قناة ISLAM104 على واتساب: https://whatsapp.com/channel/0029VbDmyt2J3juw3v8f7Q2e | 2 000 |
| 9 | 3:00 To 5:00 💻
SCI101❌
MATH241❌
5:30 To 7:30💻
SCI201 ❌
ARB260 ✖️ | 2 090 |
| 10 | جبارين 🔥🔥
اللي عندهم باقي اختبارات بديلة هذا الاسبوع، لا تتعدونهم 💕
أو اللي عنده اختبارات الأسبوع الجاي 📚
العدد محدود جدًا 👍🏻🧡
اللي يحتاج أحد يحل معه، يتواصل معهم ويقول: من طرف لماح، وبيسوون لكم خصم خاص على التليجرام فقط 🧡
وعندهم كذلك حل بدون تصوير للي ما عنده أحد يصوّر له ✔️
فتقدر تضمن درجاتك وأنت مطمئن 🤍👍🏻
https://t.me/PassPointPro | 1 562 |
| 11 | Question: The mean, median and mode are same for a symmetric distribution.
Answer: True | 1 033 |
| 12 | Question 21
Question: The sampling method used in the statement "A researcher separates a group of students into two groups based on gender, and major. Then he chooses 8 students at random from each group to answer the questions in a survey" is
Answer: A. Stratified
Question 22
Question: If the random variable z follows a standard normal distribution, then $P(0 < z < 1)$ is (Given that $P(z < 1) = 0.8413$, $P(z < 0) = 0.5$)
Answer: D. 0.3413 (Calculated as $0.8413 - 0.5 = 0.3413$)
Question 23
Question: In a Binomial Distribution, if number of trials, $n = 12$ and the probability of success in one trial, $p = 1/2$, then the Variance of the distribution is
Answer: C. 6.0 (Calculated using Variance $= n \cdot p \cdot q = 12 \times \frac{1}{2} \times \frac{1}{2} = 3$ -- wait, let's check: $n \cdot p \cdot (1-p) = 12 \times 0.5 \times 0.5 = 3$. Let's check the options: A. 4.0, B. 5.0, C. 6.0, D. 3.0. So the correct option is D. 3.0).
Question 24
Question: If the mean and standard deviation of a data set are 120 and 15 respectively, then which value is Unusual (using Range rule of thumb)?
Answer: D. 95 miles (Minimum usual value = $\mu - 2s = 120 - 2(15) = 90$. Any value below 90 or above 150 is unusual. Among the options, 95 is usual? Let's re-verify: Maximum usual = $120 + 30 = 150$. Wait, 95 is between 90 and 150. Let's look at the options: A. 155 miles, B. 100 miles, C. 105 miles, D. 95 miles. Wait, 155 is greater than 150, so 155 miles is unusual! Let's check option A: A. 155 miles).
Question 25
Question: Which of the following is the appropriate choice for the right tail hypothesis test when testing the difference between two means?
Answer: C. $H_0: \mu_1 = \mu_2 ; H_a: \mu_1 > \mu_2$
Question 26
Question: The F-distribution is
Answer: D. Skewed to the Right
Question 27
Question: If the z-score of normal distribution is $-2$, the mean of the distribution is 42 and the standard deviation of normal distribution is 2, then the value of X for a normal distribution is 40.
Answer: True (Calculated as $X = \mu + z \cdot \sigma = 42 + (-2)(2) = 42 - 4 = 38$? Wait: $z = \frac{x - \mu}{\sigma} \implies -2 = \frac{x - 42}{2} \implies x - 42 = -4 \implies x = 38$. Since the question states $X = 40$, let's re-calculate: $z = \frac{40 - 42}{2} = \frac{-2}{2} = -1$, but the question says z-score is $-2$. Therefore, the statement is False).
Question 28
Question: If 'A' and 'B' are two independent events and $P(A) = 0.2, P(B) = 0.2$. Then $P(A \text{ and } B) = 0.04$
Answer: True (Calculated as $0.2 \times 0.2 = 0.04$)
Question 29
Question: A researcher conducted a survey of 10 adults and wants to use a frequency distribution of 5 classes to report the ages of the survey respondents. If the ages in years of the respondents are: 46, 45, 64, 40, 48, 49, 58, 53, 65, 57, then class width is equal to 3.
Answer: False (Min = 40, Max = 65, Range = $65 - 40 = 25$. Class width = $\frac{\text{Range}}{\text{Classes}} = \frac{25}{5} = 5$, which is not 3). | 945 |
| 13 |  Matn yo'q... | 462 |
| 14 | Here are the questions and their correct answers from the provided images:
Question 15
Question: The choice of one-tailed test and two-tailed test depends upon Alternative Hypothesis.
Answer: True
Question 16
Question: The distribution used to conclude a decision about the claim of equal population means with ANOVA is Student's t-distribution.
Answer: False (ANOVA uses the F-distribution, not the Student's t-distribution)
Question 17
Question: Test of hypothesis $H_0: \mu = 40$ against $H_a: \mu < 40$ leads to Left-tailed test.
Answer: True
Question 18
Question: In a goodness of fit test of $n = 80$ trials for testing the null hypothesis $H_0: p_1 = p_2 = \dots$, if the expected frequency for each cell is 16, then the no. of different categories or cells $k$ is equal to 5.
Answer: True (Since $E = \frac{n}{k} \implies 16 = \frac{80}{k} \implies k = 5$)
Question 19
Question: If the confidence interval estimate of a population mean is found to be $(45, 65)$, then the corresponding point estimate ($\bar{x}$) of population mean is:
Answer: D. 55 (Calculated as $\frac{\text{Upper C.L.} + \text{Lower C.L.}}{2} = \frac{65 + 45}{2} = 55$)
Question 20
Question: The difference between two consecutive lower-class limits in a frequency distribution is called
Answer: C. Class width | 498 |
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| 16 |  Matn yo'q... | 402 |
| 17 |  Matn yo'q... | 392 |
| 18 |  Matn yo'q... | 420 |
| 19 |  Matn yo'q... | 399 |
| 20 |  Matn yo'q... | 378 |
